Question

If $f\left(x\right)={\cos }^{-1}\left(\frac{\sqrt{2x^2+1}}{x^2+1}\right)$, then range of $f\left(x\right)$ is

• A. $[0,\pi ]$
• B. $(0,\frac{\pi }{4}]$
• C. $(0,\frac{\pi }{3}]$
• D. $[0,\frac{\pi }{2})$

Answer 1

The correct option is D $[0,\frac{\pi }{2})$

Given,

$f\left(x\right)={\cos }^{-1}\left(\frac{\sqrt{2x^2+1}}{x^2+1}\right)$

Range of $\frac{\sqrt{2x^2+1}}{x^2+1}$

$0<f\left(x\right)\le 1$

${\cos }^{-1}\left(1\right)\le {\cos }^{-1}\left(\frac{\sqrt{2x^2+1}}{x^2+1}\right)<{\cos }^{-1}\left(0\right)$

$0\le {\cos }^{-1}\left(\frac{\sqrt{2x^2+1}}{x^2+1}\right)<\frac{\pi }{2}$

Therefore the range is,

$0\le f\left(x\right)<\frac{\pi }{2}$